probability and combinations

[thephoenix]

since these two are the most valued concept in GMAT , let me have the opputunity to start a thread for practicing it more thouroughly

i will post two selected problem and scratch my head to solve them.
guys if u find it helpful then do join me.

LET US STICK TO THE LEVEL OF GMAT

q1)A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 2520
B. 420
C. 168
D. 105
E. 90


Q2)2. In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?

A.27
B.81
C.168
D.280
E.729

PLS SHARE DIFFERENT APPROACHES

[firdaus117]

For solving both questions,we use fact that
The number of ways in which mn different items can be divided equally into m groups, each containing
n objects and the order of the groups is not important, is
(mn!)/{(n!)^m*m!}
Here for first question,
mn=8 n=2 m= 4 will give 105.OptionD
For second question,
mn=9 n=3 m=3 will give 280.OptionD
Note that in both question the order of making team or group is not important.

[harsh.champ]

For solving both questions,we use fact that
The number of ways in which mn different items can be divided equally into m groups, each containing
n objects and the order of the groups is not important, is
(mn!)/{(n!)^m*m!}
Here for first question,
mn=8 n=2 m=2 will give 105.OptionD
For second question,
mn=9 n=3 m=3 will give 280.OptionD
Note that in both question the order of making team or group is not important.

Hey firdaus,
Is it necessary to remember all such probability formulas??
I mean can't we approach this sum algebraically??
I know it will be lengthy that way,but once familiar with the algorithm,your speed can drastically improve.

[firdaus117]

Hey firdaus,
Is it necessary to remember all such probability formulas??
I mean can't we approach this sum algebraically??
I know it will be lengthy that way,but once familiar with the algorithm,your speed can drastically improve.

It will be better if you could remember those formulas as they will greatly reduce your answer time.Most of the questions that appear in GMAT are direct application of these formulas.However,I insist that you just don't remember these formulas rather try to understand them.
Let me explain first question as an example for you:
Cosider these 8 people standing in a row.Now grouping is a case of combination not permutation.Mentally picture three partition lines or rope between any two team.That is these ropes will be like
1 2 }3 4 }5 6 }7 8 where } represents the partition.
Now 8 people can be arranged in 8! ways.
But we have counted here a team twice say as 1 2 and 2 1.Similarly for other three teams.Hence we need to divide by 2!^4.
We now have 8!/2!^4.But is it our result....No....... why?Because we have also counted arrangement of 4 teams in between them .But in forming group of two,order of grouping is not important.So we need to divide it by 4! to correct that.
Now we get result
8!/(2!^4*4!)
Hope it helps.

[thephoenix]

For solving both questions,we use fact that
The number of ways in which mn different items can be divided equally into m groups, each containing
n objects and the order of the groups is not important, is
(mn!)/{(n!)^m*m!}
Here for first question,
mn=8 n=2 m=2 will give 105.OptionD
For second question,
mn=9 n=3 m=3 will give 280.OptionD
Note that in both question the order of making team or group is not important.

Hey firdaus,
Is it necessary to remember all such probability formulas??
I mean can't we approach this sum algebraically??
I know it will be lengthy that way,but once familiar with the algorithm,your speed can drastically improve.

in my knowledge
GMAT never test direct complex formulas , it test logics and concepts
for all problems there are at least two ways to solve

here for first one

first 2 can be selected from eight in 8c2
next two in 6c2
next two in 4c2
and last two in 2c2

tot=8c2*6c2*4C2*2c2

now order of four teams does not matters so
tot # ways=8c2*6c2*4C2*2c2/4!=[8!/(6!*2!) * 6!/(2!*4!) *4!/(2!*2!)*1]/4!=8!/(2!^4*4!)=(m*n)!/{n!^m * m!}=105

[thephoenix]

For solving both questions,we use fact that
The number of ways in which mn different items can be divided equally into m groups, each containing
n objects and the order of the groups is not important, is
(mn!)/{(n!)^m*m!}
Here for first question,
mn=8 n=2 m=2 will give 105.OptionD
For second question,
mn=9 n=3 m=3 will give 280.OptionD
Note that in both question the order of making team or group is not important.

firdaus
i think m=4 and n=2

[firdaus117]

For solving both questions,we use fact that
The number of ways in which mn different items can be divided equally into m groups, each containing
n objects and the order of the groups is not important, is
(mn!)/{(n!)^m*m!}
Here for first question,
mn=8 n=2 m=2 will give 105.OptionD
For second question,
mn=9 n=3 m=3 will give 280.OptionD
Note that in both question the order of making team or group is not important.

firdaus
i think m=4 and n=2

Just a typo.I have edited that post.